Roulette Bernoulli

 
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  1. Roulette Bernoulli Equation
  2. Roulette Bernoulli Game
  3. Roulette Bernoulli Games
  4. Bernoulli Roulette Watch

In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2., 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, you win $35; otherwise you lose $1. Complete parts (a) through (g) below. A)Construct a probability distribution for the. Let us first consider the very simple situation where the fluid is static—that is, v 1 = v 2 = 0. Bernoulli’s equation in that case is. P 1 + ρgh 1 = P 2 + ρgh 2. We can further simplify the equation by taking h 2 = 0 (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative.


Occupancy Probability for 38 Number Roulette Wheel


For our next problem, let us calculate the probabilty of getting all 38 numbers after spinning a roulette wheel 152 times.

(An American roulette wheel has the numbers 1 through 36 plus zero and double zero.)

STEP 1
To determine the number of all the results of spinning a 38 number roulette wheel 152 times we raise 'n' to the power of 'r':

38152
which equals
1.34 × 10240

Then in steps 2, 3, 4 and 5, we will determine how many of those 38152 spins, will contain all 38 numbers.

STEP 2
We must calculate each value of 'n' raised to the power of 'r'.
Rather than explain, this is much easier to show:

38152 = 1.34 × 10240
37152 = 2.33 × 10238
36152 = 3.61 × 10236
35152 = 4.99 × 10234
34152 = 6.09 × 10232
33152 = 6.52 × 10230
32152 = 6.06 × 10228
31152 = 4.86 × 10226
30152 = 3.33 × 10224
29152 = 1.93 × 10222
28152 = 9.29 × 10219
27152 = 3.69 × 10217
26152 = 1.19 × 10215
25152 = 3.07 × 10212
24152 = 6.20 × 10209
23152 = 9.61 × 10206
22152 = 1.12 × 10204
21152 = 9.49 × 10200
20152 = 5.71 × 10197
19152 = 2.35 × 10194
18152 = 6.33 × 10190
17152 = 1.07 × 10187
16152 = 1.06 × 10183
15152 = 5.83 × 10178
14152 = 1.63 × 10174
13152 = 2.09 × 10169
12152 = 1.09 × 10164
11152 = 1.96 × 10158
10152 = 1.00 × 10152
9152 = 1.11 × 10145
8152 = 1.86 × 10137
7152 = 2.85 × 10128
6152 = 1.90 × 10118
5152 = 1.75 × 10106
4152 = 3.26 × 1091
3152 = 3.33 × 1072
2152 = 5.71 × 1045
1152 = 1


STEP 3
Next, we calculate how many combinations can be made from 'n' objects for each value of 'n'.
Tthis is much easier to show than explain:

38 C 38 = 1
37 C 38 = 37
36 C 38 = 703
35 C 38 = 8,436
34 C 38 = 73,815
33 C 38 = 501,942
32 C 38 = 2,760,681
31 C 38 = 12,620,256
30 C 38 = 48,903,492
29 C 38 = 163,011,640
28 C 38 = 472,733,756
27 C 38 = 1,203,322,288
26 C 38 = 2,707,475,148
25 C 38 = 5,414,950,296
24 C 38 = 9,669,554,100
23 C 38 = 15,471,286,560
22 C 38 = 22,239,974,430
21 C 38 = 28,781,143,380
20 C 38 = 33,578,000,610
19 C 38 = 35,345,263,800
18 C 38 = 33,578,000,610
17 C 38 = 28,781,143,380
16 C 38 = 22,239,974,430
15 C 38 = 15,471,286,560
14 C 38 = 9,669,554,100
13 C 38 = 5,414,950,296
12 C 38 = 2,707,475,148
11 C 38 = 1,203,322,288
10 C 38 = 472,733,756
9 C 38 = 16,3011,640
8 C 38 = 48,903,492
7 C 38 = 12,620,256
6 C 38 = 2,760,681
5 C 38 = 501,942
4 C 38 = 73,815
3 C 38 = 8,436
2 C 38 = 703
1 C 38 = 38

Basically, this is saying that
38 objects can be chosen from a set of 38 in 1 way
37 objects can be chosen from a set of 38 in 37 ways
36 objects can be chosen from a set of 38 in 703 ways
.......................................................................................

2 objects can be chosen from a set of 38 in 703 ways
1 object can be chosen from a set of 38 in 38 ways


STEP 4
We then calculate the product of the first calculation of STEP 2 times the first calculation of STEP 3 and do so throughout all 38 numbers.

For example,
1.34 × 10240 × 1 = 1.34 × 10240
2.33 × 10238 × 37 = 8.84 × 10239
and so on
1.34 × 10240
8.84 × 10239
2.54 × 10239
4.21 × 10238
4.50 × 10237
3.27 × 10236
1.67 × 10235
6.14 × 10233
1.63 × 10232
3.14 × 10230
4.39 × 10228
4.44 × 10226
3.22 × 10224
1.66 × 10222
5.99 × 10219
1.49 × 10217
2.49 × 10214
2.73 × 10211
1.92 × 10208
8.30 × 10204
2.13 × 10201
3.07 × 10197
2.36 × 10193
9.02 × 10188
1.57 × 10184
1.13 × 10179
2.94 × 10173
2.36 × 10167
4.73 × 10160
1.81 × 10153
9.1 × 10144
3.60 × 10135
5.25 × 10124
8.79 × 10111
2.41 × 1096
2.81 × 1076
4.01 × 1048
38
Total
1.36 × 10240


STEP 5
Then, alternating from plus to minus, we sum the 38 terms we just calculated.

+ 1.34 × 10240
- 8.84 × 10239
+ 2.54 × 10239
- 4.21 × 10238
+ 4.50 × 10237
- 3.27 × 10236
+ 1.67 × 10235
- 6.14 × 10233
+ 1.63 × 10232
- 3.14 × 10230
+ 4.39 × 10228
- 4.44 × 10226
+ 3.22 × 10224
- 1.66 × 10222
+ 5.99 × 10219
- 1.49 × 10217
+ 2.49 × 10214
- 2.73 × 10211
+ 1.92 × 10208
- 8.30 × 10204
+ 2.13 × 10201
- 3.07 × 10197
+ 2.36 × 10193
- 9.02 × 10188
+ 1.57 × 10184
- 1.13 × 10179
+ 2.94 × 10173
- 2.36 × 10167
+ 4.73 × 10160
- 1.81 × 10153
+ 9.10 × 10144
- 3.60 × 10135
+ 5.25 × 10124
- 8.79 × 10111
+ 2.41 × 1096
- 2.81 × 1076
+ 4.01 × 1048
- 38
Total
+ 6.72 × 10239
6.72 × 10239 equals the total number of ways a 38 number
roulette wheel will show all 38 numbers after 152 spins.

STEP 6
So, if we take the number
1.36 × 10240 (all results of spinning a 38 number roulette wheel 152 times)
and divide it by
6.72 × 10239 (all results of all 38 numbers appearing after 152 spins),
we get the probability of rolling all 38 numbers appearing after 152 spins.
Probability = 6.72 × 10239 ÷ 1.34 × 10240 = 0.501599170962349

Basically, you would have to spin a roulette wheel at least 152 times in order to have a better than 50 / 50 chance of spinning all 152 numbers.


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Bernoulli distribution Random number distribution that produces bool values according to a Bernoulli distribution, which is described by the following probability mass function: Where the probability of true is p and the probability of false is (1-p). Palace Casino is betting at a Roulette table. He is following a gambling strategy that is often used by prudent gamblers. He has dedicated a capital of $200 to this session with the plan of not winning more than $20. He invariably bets $1 on RED at each spin and plans to do continue playing. Bernoulli trials form one of the simplest yet still most important of all random processes. Ifp= 0 then the gambler always loses and ifp= 1 then the gambler always wins. These trivial cases are not interesting, so we will usually assume that 0.

A Bernoulli Distribution is the probability distribution of a random variable which takes the value 1 with probability p and value 0 with probability 1 – p, i.e.

$$
begin{cases}
1-p & text{for} k=0
p & text{for} k=1
end{cases}$$

We will use the example of left-handedness. Approximately 10% of the population are left-handed (p=0.1).

We want to know, out of a random sample of 10 people, what is the probability of 3 of these 10 people being left handed?

We assign a 1 to each person if they are left handed and 0 otherwise:

Bernoulli roulette watch
  • $P(X=1) = 0.1$
  • $P(X=0) = 0.9$

A Binomial distribution is derived from the Bernoulli distribution.

We’ll start with the simpler problem:

What is the probability of the first 3 people we pick being left-handed, followed by 7 people being right-handed?

This is just $ 0.1 ^3 times 0.9 ^7$

What if we wanted the last 3 people to be left-handed?

This is just $0.9^7 times 0.1^3$, the same answer.

In fact, no matter how we arrange the 3 people, we will always end up with the same probability ($ 4.7 times 10^{-4} $).

So we have to add up all the ways we can arrange the 3 people being picked.

There are $10!$ ways to arrange 10 people and there are $3!$ ways to arrange the 3 people that are picked and $7!$ ways to arrange the 7 people that aren’t picked.

Roulette Bernoulli

This is given as:
$$dfrac{10!}{3! 7!}$$

Or more commonly, “10 choose 3”. The “n choose k” notation is written as:
$$
begin{equation*}
binom{n}{k}
end{equation*} = dfrac{n!}{k! (n-k)!}
$$

We can now caclulate the probability that there are 3 left-handed people in a random selection of 10 people as:

$$
P(X=3) = begin{equation*}
binom{10}{3}
end{equation*} (0.1)^3 (0.9)^7
$$

This will generalise such that:

$$
P(X=k) = begin{equation*}
binom{n}{k}
end{equation*} (p)^k (1-p)^{n-k}
$$

Roulette Bernoulli Equation

Scipy’s stats package has a binomial package that can be used to calculate these probabilities:

We can use this function to calculate what the probability of 3 or fewer people being left-handed from a selection of 10 people.

$$
P(X leq 3) = sum_{i=0}^{3} begin{equation*}
binom{10}{i}
end{equation*} (0.1)^i (0.9)^{n-i}
$$

Or we could plot our probability results for each value up to all 10 people being left-handed:

We can see there is almost negligible chance of getting more than 6 left-handed people in a random group of 10 people.

Roulette

On an American roulette wheel there are 38 squares:

  • 18 black
  • 18 red
  • 2 green

We bet on black 10 times in a row, what are the chances of winning more than half of these?

$$
P(X gt 5) = sum_{i=6}^{10} begin{equation*}
binom{10}{i}
end{equation*} bigg(dfrac{18}{38}bigg)^i bigg(1-dfrac{18}{38}bigg)^{n-i}
$$

A Poisson distribution is a limiting version of the binomial distribution, where $n$ becomes large and $np$ approaches some value $lambda$, which is the mean value.

Roulette Bernoulli Game

The Poisson distribution can be used for the number of events in other specified intervals such as distance, area or volume. Examples that may follow a Poisson include the number of phone calls received by a call center per hour and the number of decay events per second from a radioactive source.

It is calculated as:

$ P(k) = e^{-lambda} dfrac{lambda^k}{k!} $

The average number of goals in a World Cup football match is 2.5.

Roulette Bernoulli Games

We would like to know the probability of 4 goals in a match.

Bernoulli Roulette Watch

Again, scipy has in-built functions for calculating this and we can use this to calculate the probability of any number of goals in a World Cup match.

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